Find three consecutive positive even integers such that the product of the second and third integers is twenty more than ten times the first integer
You should come up with the following notation for the three consecutive even integers, such that:
`n, n+2, n+4`
The problem provides an information that relates the integers, such that:
`(n+2)(n+4) = 20 + 10n`
`n^2 + 4n + 2n + 8 - 10n - 20 = 0 => n^2 - 4n - 12 = 0`
Using quadratic formula yields:
`n_(1,2) = (4+-sqrt(16 + 48))/2`
`n_(1,2) = (4+-sqrt64)/2 => n_(1,2) = (4+-8)/2`
`n_1 = 6; n_2 = -2`
Since the problem provides the information that the integers are positive, hence, only `n = 6` fulfills the condition.
Hence, evaluating the three consecutive even positive integers, yields 6, 8, 10.